Problem: Simplify; express your answer in exponential form. Assume $n\neq 0, x\neq 0$. $\dfrac{{(n^{3})^{-2}}}{{(n^{-4}x^{-5})^{4}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${n^{3}}$ to the exponent ${-2}$ . Now ${3 \times -2 = -6}$ , so ${(n^{3})^{-2} = n^{-6}}$ In the denominator, we can use the distributive property of exponents. ${(n^{-4}x^{-5})^{4} = (n^{-4})^{4}(x^{-5})^{4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(n^{3})^{-2}}}{{(n^{-4}x^{-5})^{4}}} = \dfrac{{n^{-6}}}{{n^{-16}x^{-20}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{-6}}}{{n^{-16}x^{-20}}} = \dfrac{{n^{-6}}}{{n^{-16}}} \cdot \dfrac{{1}}{{x^{-20}}} = n^{{-6} - {(-16)}} \cdot x^{- {(-20)}} = n^{10}x^{20}$.